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3241004551 [841]
2 years ago
9

Draw the schematics for the following two battery connections. Can you explain the value of the output voltage in the series con

nection
Engineering
1 answer:
matrenka [14]2 years ago
5 0

if two 1.5 ya alt batteries are connected to head to the tail the voltage is 3.0 volt it is the because the battery is insidious reduce a voltage equal to number of battery is multiplied by the voltage of individual

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The three sub regions of South America are the Andes Mountains, the Amazon Rainforest, and the Eastern Highlands. The Atacama De
Leto [7]

Answer:

<:

Explanation:

8 0
3 years ago
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What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit
jeka94

Answer:

a. 0.28

Explanation:

Given that

porosity =30%

hydraulic gradient = 0.0014

hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

v=\dfrac{K}{n_e}\dfrac{dh}{dl}

v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s

v=3.22\times 10^{-6}\ m/s

The velocity in m/d      ( 1 m/s =86400 m/d)

v= 0.27 m/d

So the nearest answer is 'a'.

a. 0.28

4 0
3 years ago
What line separates two lanes traveling in the same direction
soldier1979 [14.2K]

Answer:

White lane lines separate lanes of traffic moving in the same direction. (UK)

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2 years ago
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What are the advantages of triggering circuit<br> for thyrestors?
sergejj [24]

Answer:

For most applications, it is simple, dependable, efficient, and straightforward to apply - a simple trigger signal may be provided, with appropriate processing if necessary. This implies that an appropriate trigger signal may be generated using other electrical circuits and then applied to the SCR.

Explanation:

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3 years ago
If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2}  } , if n_{i}=1.5x10^{10}cm^{-3}  \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15}  }{(1.5x10^{10})^{2}  } )=0.83V

w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15}  } } =3.35x10^{-5} cm=0.335um

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

7 0
3 years ago
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